SNOAA73 May 2021 LMP2012QML-SP , LMP7704-SP , TPS7A4501-SP

A 4–20-mA V-to-I current
transmitter takes a differential voltage as the input and outputs a current with a
linear relationship from input to output. This linear relationship depends on the
common-mode range of the op amp and the headroom available for the transistor and
LDO. Furthermore, a current transmitter designed with a 2-wire loop requires that
the quiescent current drawn from all components on the loop path consume less than
the minimum output current I_{out_min} – 4 mA in this design. This includes
the op amp, LDO, the current through the R_{3} and R_{5} branch, and
any other peripherals that are on the same supply loop. As long as this requirement
is met, the op amp and transistor compensate to supply the remaining current to the
loop to output the correct current to the load resistor.

The output current I_{out} is
the sum of the currents through R_{3} and R_{4}, I_{3}, and
I_{4}, respectively:

I_{3} is set by R_{1}
and R_{2}, along with the input voltage V_{in} and regulated supply
V_{reg}. Under ideal operation, the non-inverting terminal of the op amp
has a virtual short to the local ground, I_{ret}. Notice also that
V_{reg} and V_{in} are referenced to I_{ret}.
I_{3} is then the following:

I_{4} is the sum of the local
ground return current I_{ret} and the compensating current through the
transistor T_{1}.

For correct operation of the op amp,
the voltages across R_{3} and R_{4} must be equal, so we have
V_{3} = V_{4} and it
follows that:

The previous equation is rewritten as:

Combining the first equation, second equation, and previous equation, the transfer function for the circuit is derived as:

It is apparent, then, that the current
through R_{3} that is set by R_{1}, R_{2}, V_{in},
and V_{reg} is amplified by the factor

1 +
R_{3} / R_{4}.